How do I use rational zero theorem and descarte's rule of signs to find all zeros of these given functions

I don't get what I'm suppose to do!

1. 3x^3-8x^2-8x+8=0
2. f(x)= x^4-4x^3-x^2+14x+10
3. 3x^4-11x^3-3x^2-6x+8 = 0

1) rational roots theorem states that all possible roots are of the form:

+/- (any factor of 8)/(any factor of 3) the 8 is the constant monomial, the 3 is the first coef

these would be

+/- 8/3, +/- 8/1, +/- 4/3, +/- 4/1, +/- 2/3, +/- 2/1, +/- 1/3, +/- 1/1


so for number 2, the possible roots would be

+/- (any factor of 10)/(any factor of 1)


You would use synthetic division to check these possibilities one by one to find a zero. The quotient from synthetic division should then be factored/quad formula if it is a quadratic, or you could repeat the whole process if it is of degree 3 or more.

for number 1, try synthetic division with the possible zeros until you find one that gives no remainder

2/3 works out:

2/3 in the box with 3 -8 -8 8
2 -4 -8
3 -6 -12 0 notice the 0 remainder

thus #1 equals (x-2/3)(3x^2 - 6x - 12) now I would factor out 3 from the second set of parenthesis

(x-2/3)(3)(x^2 - 2x - 4) and use the quadratic formula on the x^2 - 2x - 4 to find the other zeros

the 3 that was factored out is irrelevant to the zeros

the 3 zeros are: 2/3, 1+/- sqrt(5)


for #2 and #3 you will need to do the synthetic division step 2 times as the quotient from synthetic division will be a cubic polynomial - you should use the rational zeros theorem again with this step



descarte's rule of signs isn't all that helpful with these

here is a link on it:

http://www.purplemath.com/modules/drofsi…